I'm trying to isolate x from this equation:

$$0.6^{x+2}-x-2 = 0$$

I know I need to use logarithms somehow and then I need to find a derivative of isolated x. Tried to use wolfram suggestions with no luck. Hope to find suggestions/help from mathematics community. Thanks.

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    $\begingroup$ There is no nice way to solve for $x$ using elementary functions. It is possible to approximate $x$ numerically, and we can find $x$ exactly using advanced functions like the Lambert $W$ function. $\endgroup$ – vadim123 Dec 23 '14 at 17:11

$$ y = x+2 $$ results in $$ 0.6^y -y = \mathrm{e}^{y\ln(0.6)} -y = 0 $$ thus $$ y\mathrm{e}^{-y\ln(0.6)} = 1 $$ let $z = -y\ln(0.6)$ then $$ \frac{-1}{\ln(0.6)}ze^z = 1 $$ or $$ z\mathrm{e}^z = -\ln(0.6) = \ln(5/3) $$ or $$ z = W(\ln(5/3)) $$ where $W$ is the lambert function or $$ x = \frac{1}{\ln(5/3)}W(\ln(5/3))-2 $$ but this is dependent if you consider the Lambert as a solution..

  • $\begingroup$ It's somewhat circular, as that function is defined to make the equality hold. It's as if I say what $\pi$, oh well it equals $\pi$. Calculating $W$ is the same problem he was asking in the first place. $\endgroup$ – Matthew Levy Dec 23 '14 at 17:25
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    $\begingroup$ I know right? But it does look pretty :). I will convert to a community post as it is not really a solution in this sense. $\endgroup$ – Chinny84 Dec 23 '14 at 17:27

Hint: the solution $x_0$ is the $x$ intercept of the graphs of $y = 0.6^{x+2}$ and $y = x+2$, and using W.A I got $x_0 \approx -1.30047$.

  • $\begingroup$ I don't need a solution, I need isolated x. $\endgroup$ – Neone Dec 23 '14 at 17:13
  • $\begingroup$ You can't isolate $x$. $\endgroup$ – DeepSea Dec 23 '14 at 17:14

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